# Almost all Regular Graphs are Hamiltonian

## Almost all Regular Graphs are Hamiltonian

Europ. J. Combinatorics (1983) 4, 97-106 Almost all Regular Graphs are Hamiltonian BELA BOLLOBAS 0. INTRODUCTION In what models of random graphs i...

Europ. J. Combinatorics (1983) 4, 97-106

Almost all Regular Graphs are Hamiltonian BELA BOLLOBAS

0.

INTRODUCTION

In what models of random graphs is it true that almost every graph is Hamiltonian? This is one of the important questions that Erdos and Renyi proposed but did not answer in their fundamental paper [8] on the evolution of random graphs. More specifically, for what functions M (n) is it true that almost every labelled graph with n vertices and M (n) edges contains a Hamilton cycle? After several preliminary results due, among others, to Moon [15], Wright [19], Perepelica [16], and Koml6s and Szemeredi [10], a break­ through was achieved by P6sa [17] and Korshunov [13]. They proved that for some constant c almost every labelled graph with n vertices and at least en log n edges is Hamiltonian. Subsequently Angluin and Valiant [1] found a fast probabilistic algorithm for constructing a Hamilton cycle. It is easily seen that if almost every graph with n vertices and M (n) edges has a Hamilton cycle then

n

M(n) ;32{log n +log log n +w(n)},

(0)

where w(n)~oo. Recently Korshunov [14] and Koml6s and Szemeredi [11] proved that this simple necessary condition is, in fact, sufficient. Our aim in this paper is to study Hamilton cycle in random regular graphs. It was conjectured in [5] that almost every four regular graph is Hamiltonian. In view of the results above this is a fairly surprising conjecture since such a graph of order n has only 2n edges. Though we shall not prove this conjecture, we s.hall show that if k is a sufficiently large constant then almost every k-regular graph is Hamiltonian. We shall also give a short proof of the fact that condition (0) is sufficient to imply the existence of a Hamilton cycle in almost every graph. Recently Fenner and Frieze [9] and Koml6s and Szemeredi [12] solved a problem closely related to the Hamilton cycle problem in random regular graphs. Given 1 ,;;: k ,;;: n - 1 and n labelled vertices, for each vertex x join x randomly to k of the other vertices. Let Gk-out be the random graph formed by the union of these edges. Notice that Gk-out has at most kn edges (it can have as few as kn/2) but every vertex has degree at least k. Fenner and Frieze and Koml6s and Szemeredi proved that if k is sufficiently large then almost every Gk-out is Hamiltonian. (It is very likely that this assertion holds for all Gk-out ;3 2.) The proof of our main theorem is rather similar to the proofs of the assertion above and somewhat reminiscent to the proofs in P6sa [17] and Koml6s and Szemeredi [11]. All these proofs rely on a lemma of P6sa [17] and so will ours. The systematic study of random regular graphs has begun only recently. The reason for this is that an asymptotic formula for the number of labelled regular graphs was found only in 1978 by Bender and Canfield [2]. Even more rece~tly, in [4], a model was constructed for the set of regular graphs that makes it fairly easy to study random regular graphs. This model is the other cornerstone of the proof of our main theorem. For the terminology and basic results in probabilistic graph theory the reader is referred to [3]. In particular, we write F(U) for the set of vertices joined to at least one vertex in a set U. 97

0195-6698/83/020096+10 \$02.00/0

98

B. Bollobds

1.

THREE LEMMAS

Let us start with three lemmas we shall need in the sequel. The first of these is a variant of the lemma of P6sa mentioned above. LEMMA 1. Suppose h, u EN and a graph B is such that its longest path has length h but it does not contain a cycle of length h + 1. Suppose furthermore that for every set U with IUl < u we have IU ur(U)I ~3IUI.

Then there are u distinct vertices Yt. y 2 , ••• , Yu and u not necessarily distinct sets Yt. Y2, ... , Yu such that fori= 1, 2, ... , u we have y;e Y;, IYd ~ u and no edge of B joins y; to Y;. Furthermore, the addition of any of they;- Y; edges creates a cycle of length h + 1.

PROOF. Let p = XoX1 ... Xh be a longest path. Then r(xh) c (XI, •.• ' Xh-1) since other­ wise P could not be extended to a longer path. Now if xh is joined to x; then P' = XoX1 . .. x;xhxh-1 ... x;+; is also a longest xo-path, that is a longest path beginning at xo. P' is called a simple transform of the x 0 -path P. The result of a sequence of simple transforms is called a transform. By a lemma of P6sa [17] (see also [3, p. 79, Theorem 15]) if U is the set of endvertices (other than x 0 ) of the transforms of P then IU ur(U)I ~3IUI-1,

and by joining x 0 to a vertex of U either we make the graph Hamiltonian or we increase the length of its longest path. Hence by our assumption IUl ~ u, say {y1y2, ... , Yu} cU.

For each y; there is a y;-path P; of maximal length h, which is a transform of Pending at y;. Let Y; be the set of endvertices (other than x 0 ) of transforms of they-path P;. As before, IYd ~ u, y;e Y;, no edge joins y; to Y; and the addition of a y;- Y; edge creates a cycle of length h + 1. To state the next two lemmas some definitions are needed. Let k ~ 107 be fixed. A graph G of order n is said to be dispersed if (a) any t ~ n/(250e) vertices span fewer than 3t/2 edges, (b) any t ~ n/16 vertices span fewer than 99k t/400 edges. A set of 3n edges of a graph H of order n is said to be dispersed if at most n/400 vertices are not incident with any of the 3n edges, at most (n/720)(606/ k )k1101 vertices are incident with at least k/101 of the 3n edges. Let '§ be the set of k -regular graphs with a fixed set V of n distinguishable vertices. It will always assume that n > k and kn is even, so that '§ is not empty. '§is viewed as a probability space in which all points have the same probability. Almost all k-regular graphs are said to have a property (lJ> if the probability that G E '§has (lJ> tends to 1 as n ~ oo. LEMMA

2.

Almost every k-regular graph is dispersed.

PROOF. It is known (see e.g. [6]) that if e tk - 1>14

k2(llk)M ­ 4n

·

Almost all regular graphs are Hamiltonian

99

This inequality implies that the expected number of t-sets of vertices spanning at least

M edges is at most

(1)

Now if M = f3t/2l and t ~ fn/(2SOe)l then the right-hand side of inequality (1) is at most

If 1 ~ t ~ n 112 then 115 Ot ~c1n-

(2)

112 for some C1 > 0 and if n ~ t ~ fn/ (250e)l then Qt~c22-tl2,

(3)

where c 2 > 0 is a constant. Finally, if M = ficfoktl and t ~ n/ 16 then the right-hand side of inequality (1) is at most = k2 (99/400)(k-1)1(500) (99/400)kl I e 33 Rt e n

(!_)

(4) for some constant c3 > 0. The probability that a k-regular graph is not dispersed is at most L:Ot+L:Rr.

where the first summation is up to fn/(2soe)l and the second summation is from 115 n/(2Soen to Ln/16J. By relationships (2), (3) and (4) this sum is O(n - ).

r

LEMMA 3. Let H be a graph of order n with size M, maximal degree at most k- 6 and minimal degree at least k - 8. Then H contains at least

(1- en ­

113

)(:!)

dispersed sets of 3n edges, where c = c (k) depends only on k.

Consider the (~) sets of 3n edges of H as a probability space n. For 0 ~ d ~ k - 6 and E En denote by Xd (E) the number of vertices of H incident with precisely d edges of E. Then Xd =Xd(E) is a random variable on n. Let V(H) = {xh x 2 , ••• , Xn} and d; = d(x;), i = 1, 2, ... , n. Then by assumption PROOF.

k - 8 ~ d;

~

k - 6 for every i

and

Clearly

(5)

100

B. Bollobds

and

E(X~) =E(Xd)+2 I' (d;; 1)(di; 1)(M ~:i~2d~ + 1) I(~)

2

+ I' (d; d

-1)(did -1 -1)(M3n-d;- 2d-di++11)/(M) -1 3n

+2I"(d)(:i)(~~~i2~di)/(~),

(6)

where I' denotes summation over all pairs (i, j) with i < j and x;xi E E (H) and I" denotes summation over the pairs (i, j) with i =E(xd)+2

Ii +2 n+2 I",

where the terms denote the corresponding terms in equation (6). Notice first that

where c 1 = c 1 (k) depends only on k. Furthermore, by expanding the square of expression (5) and comparing the terms with the terms in I", we find that 2 I"/E(Xd)

2

1 -d·)(M -d·) M -d· -d·)(M)/(M 3n 3n-d 3n-J'

~~~x ( 3n-,2d

where the maximum is taken over all pairs (i, j) with d that a quotient on the right is at most

~

d;

~ di.

Crude calculations show

M-d-+1 )d' ,.::: 1 +­ 3k ------'-'-( M-d--d-+1 ~ n' ,

I

I

so

Consequently,

3)

E(X 2d)~E(Xd)+2ctn + ( 1 +;;- E(Xd) 2 2

~E(Xd) +c2n,

and therefore

P(iXd- E(Xd)i;;;. n 3) ~ c2n -,_ 2

1

(7)

It is now shown that expression (7) implies the assertion of Lemma 3. Indeed, X 0 is the number of vertices not incident with any of the edges of E, and

E(Xo)=

f

;~1

(M-d;)/(M)~n(M-k+8)/(M) 3n

3n

~ n( 1 - k~ 8) " ~ nexp { -6 kk 3

3n

=8}5 .

3n

Almost all regular graphs are Hamiltonian

101

Since exp[ -6(k- 8)/(k- 5)],; e - 5 .9999 < 1/400, by (7)

(8) for some c 3 > 0 and every n. For EEil the number of vertices incident with at least k 0 = rk/1011 edges of E is Y = L . . ko ~·. Notice that by equation (5)

(

E(Y) =E L

j.2
,;n

xi) = j.2
-k +8)/(M) . . (k-6)(M 3n 3n -1 1 6 )i . ---­ L (k -6)(

E(Xj):s;n L 0

j;;.k 0

. M-3n3n )i :s;n L (k-6)( 1

i.,..ko

1

j.2
k -14

Combining this equality with expression (7) gives that

1 606e) P( y;;;.non(--k-

k/

101

)

'

:s;c4n-',

(9)

for some c 4 > 0 and every n. Inequalities (8) and (9) imply Lemma 3. 2. THE MAIN THEOREM Before stating our main result let us describe the model of k -regular graphs given in [4]. Let k;;;. 3 be fixed and suppose n ~ oo such that kn is even. Let W = U~= 1 Wi be a fixed set of kn elements partitioned into n sets of I<; elements each. A configuration is a partition of W into m = kn/2 pairs of elements. Denote by il the set of configurations and set N(m) = lill. Notice that N(m) = 2-m(2m)!/m!. Given a configuration C denote by c/J (C) the graph with vertex set V = {1, 2, ... , n} in which ij is an edge iff C contains a pair with one element in Wi and the other in Wi. Every graph c{J(C) has maximal degree k. Furthermore, if G is a k-regular graph with 1 vertex set V then there are exactly (k !t configurations in c/J - ( G). Thus if, as earlier, '§ denotes the set of k -regular graphs with vertex set V then '§ c c{J (il) and

The set c{J - ('§) = il* consists of those configurations in which at most one pair is2contained 1 14 in the sets of the form Wiultj, ii'j. It is shown in [4] that lil*l-e-(k - l lill= 2 e-(k'- 1 l14N(m) so if O
cN(m),; (k !tl'§l :s;N(m). THEOREM 4.

(10)

7 Let k ;;;.10 be fixed. Then almost every k-regular graph is Hamiltonian.

PROOF. Denote by '§0 the set of graphs in '§ that are dispersed, k-connected and are not Hamiltonian. Since almost every k-regular graph is k-connected (see [5]) and by Lemma 2 almost every k-regular graph is dispersed, the theorem follows if we show that almost no graph belongs to '§0 , that is l'§ol/1'§1 ~ 0.

102

B. Bollobtis

Let G E C§0 • Since G is k-connected and regular, it contains a 6-factor Fa (see [7]). Furthermore, as G is connected and not Hamiltonian, its longest path contains more vertices than its longest cycle. Let Pa be a longest path in G. Consider all red-blue colourings of the edges of G in which there are exactly 3n red edges, the edges of Fa and Pa are blue, and the 3n red edges are dispersed. We call these colourings admissible. Notice that admissible colourings correspond to dispersed sets of 3n edges of the graph G-E(FauPa). By Lemma 3, if n is sufficiently large, every G E C§o has at least

admissible colourings. Given an admissible colouring c of a graph G E C§0 , denote by Gc the subgraph of G spanned by the edges that are blue in the colouring c. LEMMA 5.

Let G

E C§0

and let

c

be an admissible colouring of G. Then the graph

B = Gc has the property that if U c V(G) and lUI:;;;; n/48 then

IUuFB(U)I>3IUI. Suppose there is a set U contradicting the assertion, say u 1 =I Ul:;;;; n/ 48 and u2 = IU u rB(U)I:;;;; 3IUI = 3ul. Denote by p the number of edges of G joining the vertices of U, by q the number of U- (F8 ( U)- U) edges of G and by r the number of other edges of G incident with the vertices of U. Notice that these last r edges are all red and PROOF.

r+2p +q = kiUI = ku1.

(11)

2p +q ;;,;6IUI = 6u1.

(12)

Furthermore, since Fa cB, Set {3 = (2p +q)/(4u 1). Then either (13)

or else (14) since p<{3u1 andp+q<3{3ul

would imply 2p +q <4{3u 1 = 2p +q.

From inequalities (13) and (14) one can conclude that there is a set T c V(G), such that Uc T, ITI:s;3IUI and T spans at least f31TI edges of the graph G. Now if lUI:;;; 3 n/(750e) then ITI:;;;; n/(250k ) and by expression (12) {3;;,; 3/2, that is T spans at least (3/2)ITI edges. This is impossible, since the graph G is dispersed. Suppose now that and so

Almost all regular graphs are Hamiltonian

By Lemma 3 r

~ u 1101 + 720 (-kk

kn

1650) k/

101

103

k

< u 1100'

so equation (11) implies that {3 =

2p +q

99

4u-;- ~ 400 k.

This again contradicts the fact that G is dispersed since it implies the existence of a set T with at most n/16 vertices that spans at least (99/400)kiTI edges. We now continue the proof of Theorem 4. A partition of 2m -6n = kn -6n elements of W=U7=1 W; into m-3n disjoint pairs will be called a subconfiguration. Set C€ 0 = ¢ -\~o) so that IC€ol = (k !tl~ol· If c is an admissible colouring of G E ~o and C E ¢-\G) then c induces a colouring of C; by cc is denoted the subconfiguration formed by the blue pairs of C. Notice that one obtains at least A(n)IC€ol coloured configurations. Suppose P is a configuration that can arise as a blue subconfiguration, that is P = cc for some admissible colouring of ¢ (C) E ~0 • The number of coloured configurations are now estimated for which the blue subconfiguration is exactly P. Set u = fn/481. Suppose B is a graph that can arise as the blue subgraph in an admissible colouring of a graph in ~o. Lemmas 1 and 5 imply that B contains distinct vertices y 1 y2 , ••• , Yu and not necessarily distinct sets Yt. Y2 , ••• , Yu such that IYd = u and if B = Gc for some G E ~0 and admissible colouring c of G then no edge of G joins y; toY;. Since u-n/400~In/55l=v, at least v of the vertices y; are incident with red edges and for every i at least v vertices in Y; are incident with red edges. This implies that for our P we can find distinct elements Zt. z 2 , ••• , Zv and not necessarily distinct sets Zt. Z2, ... , Zv, such that IZd = v, no pair of P contains an element of {zt. ... , zv}u Z 1 uZ2 u · · · uZv, and if P = Cc then in the configuration C no element Z; is paired with an element of Z;. How can one find a configuration C for which cc = P? In P there are kn - 6n elements of W partitioned into pairs, so to obtain such a C one has to partition the remaining 6n elements into pairs. Let the pairs containing Zt. z 2 , ••• , Zv be chosen consecutively. There are at most 6n - v choices for the pair containing z 1 since z 1 cannot be paired with an element in Z 1 • Suppose l pairs have been chosen. Let j be the smallest index for which zi is not in a pair. Then there are at most 6n - v choices for the pair containing zi; choose one of them. Continuing in this way we find that each of the first w = lv/2J pairs can be chosen in at most 6n - v ways. The remaining 3n - w pairs of a C with Cc = P can be chosen in at most N (3n - w) ways. Hence there are at most (6n -v)wN(3n -w)

configurations with Cc = P. Notice that (6 n -v)wN( 3 n -w)/N( 3 n) = (6n -vt2w3n(3n -1) · · · (3n -w + 1) 6n(6n -1) · · · (6n -2w + 1) (6n-v)w (6n -1)(6n -3) · · · (6n -2w + 1) ~exp[

J

1 Lw -v -2i - -+­ i=16n-2i+1

104

B. Bollobds

~

~ v-2i+1] [ w(v-w)] exp [ - '-= exp ­ i=l 6n 6n

< 2 exp[

7 ~~ 0 ].

Therefore for every subconfiguration P at most (15) configurations C satisfy Cc = P for some admissible colouring c. We are ready to put together all the information. The number of subconfigurations containing m - 3n pairs is

N(m)(;,:)/ N(3n), since every configuration contains(;,:) subconfigurations and every subconfiguration is contained in N(3n) configurations. Consequently by expression (15) there are at most 2 exp[

7 ~~ 0 ]N(m)(;,:)

coloured configurations. On the other hand, there are at least 1

A(n)l~ol=:z

(m 3-4n) n (k!) nl CSo I

coloured configurations. Therefore

Since m ( 3n

)/(m-3n4n) < ( m-7n m - 3n)

3n

< exp[ 12n z

m-7n

J< exp(25n 10-7)

and 25·72600 < 10 7 ,

ICSol = o(N(m)(k !)-n). By inequality (10) this is equivalent to

ICSoi/ICSI ~ 0 as n ~ oo. The proof of Theorem 4 is complete. 3.

THE MODEL

CS(n, M)

To conclude this paper a proof of the exact condition (0) is sketched for the existence of a Hamilton cycle in a random graph of order n with M = M (n) edges. Though the method of proof of Theorem 4 can easily be adapted to give a proof of the sufficiency of (0), a slightly different approach will. be chosen here, where the proof is based on a theorem of de la Vega [18]. To be precise the result needed is the following: if

Almost all regular graphs are Hamiltonian

105

M 0 = M 0 (n);;;.: 1/2n log n then almost every random graph GM E C§(n, M) contains a path of length at least n -(en/log n ), where e is a positive constant. This is proved in [18] but it is not stated explicitly. THEOREM 6. If M=M(n);;;.:nj2[logn+loglogn+w(n)], where w(n)-+oo, then almost every graph of order n and size M is Hamiltonian. We shall only sketch a proof and leave the details to the reader. Setk = len/log n J, where e is theconstantappearingabove. Putm = f32log n l andM0 =M -km. ThenM0 =M0 (n) satisfies the same inequality as M (n ), with w (n) replaced by w 0 (n) = w (n)- 2km/n -+ oo. It is easily seen that almost every GMo is connected and has minimum degree at least 2. Furthermore, almost every GMo is such that whenever U is a set of at most n/4 vertices, Jr(U) u Ul;;;.: 3IUJ. Let G 0 = GMo be a graph having the three properties above. Add m edges at random to obtain Gt. then add m edges to G 1 at random to obtain G 2 , and so on. The theorem follows if we show that with probability tending to 1 the graph Gk is Hamiltonian. Suppose G; is not Hamiltonian. As in the proof of Theorem 4, there are at least (1/2)(n/4) 2 edges, such that the addition of any of these edges to G; either makes the new graph Hamiltonian or it increases the length of a longest path. The probability that none of m edges is chosen from a set of at least (h(2)(n/4) 2 edges is at most

Hence if Gh is not Hamiltonian and /(G) denotes the length of a longest path of G then with probability at least 1- kn - 2 we have n- k ~ l(G 0 ) < l(G 1) < · · · < l(Gh) ~ n -1. Since this is impossible, the probability that Gh is Hamiltonian is at least 1 - kn - 2 = 1-o(1). This completes the proof. REFERENCES 1. D. Angluin and L. G. Valiant, Fast probabilistic algorithms for Hamiltonian circuits and matchings, J. Computer and System Sciences 18 (1979), 155-193. 2. E. A. Bender and E. R. Canfield, The asymptotic number of labelled graphs with given degree sequences, J. Combin. Theory Ser. (A) 24 (1978), 296-307. 3. B. Bollobas, Graph Theory-An Introductory Course, Graduate Texts in Mathematics, Springer, New York, 1979. · 4. B. Bollobas, A probabilistic proof of an asymptotic formula for the number of labelled regular graphs, Europ. J. Combinatorics 1 (1980), 311-316. 5. B. Bollobas, Random Graphs, in Combinatorics (H.N.V. Temperley, ed.), London Math. Soc. Lecture Notes No. 52, Cambridge University Press, Cambridge, 1981, pp 80-102. 6. B. Bollobas, The asymptotic number of regular unlabelled graphs, j London Math. Soc. (2) 26 (1982), 201-206. 7. B. Bollobas and N. C. Wormald, Regular factors of regular graphs, to appear. 8. P. Erdos and A. Renyi, On the evolution of random graphs, Pub/. Math. Inst. Hungar. Acad. Sci. S (1960), 17-61. 9. T. I. Fenner and A.M. Frieze, On the existence of Hamiltonian cycles in a class of random graphs, to appear. 10. J. Koml6s and E. Szemeredi, Hamiltonian cycles in random graphs, in Infinite and Finite Sets: Colloq. Math. Soc. Janos Bolyai 10 (A. Hajnal, R. Rado and V. T. S6s, eds.), North-Holland, Amsterdam, 1975, pp 1003-1011. 11. J. Koml6s and E. Szemeredi, The exact probability distribution of Hamiltonian cycles in random graphs, to appear. 12. J. Koml6s and E. Szemeredi, to appear. 13. A. D. Korshunov, Solution of a problem of Erdos and Renyi on Hamilton cycles in nonoriented graphs, Soviet Math. Doklady 17 (1976), 760-764.

106

B. Bollobds

14. A. D. Korshunov, A solution of a problem of P. Erdos and A. Renyi about Hamilton cycles in non-oriented graphs (in Russian), Metody Diskr. Anal. v Teoriy Upr. Syst., Sbornik Trudov Novosibirsk 31 (1977), 17-56. 15. J. W. Moon, Almost all graphs have a spanning cycle, Canad. Math. Bull. 15 (1972), 39-41. 16. V. A. Perepelica, On two problems from the theory of graphs, Soviet Mat. Dokl. 11 (1970), 1376-1379. 17. L. P6sa, Hamiltonian circuits in random graphs, Discrete Math. 14 (1976), 359-364. 18. W. F. de Ia Vega, Long paths in random graphs, to appear. 19. E. M. Wright, For how many edges is a graph almost certainly Hamiltonian? J. London Math. Soc. 8 (2), (1974), 44-48. Received 26 February 1982 BELA BOLLOBAS

Louisiana State University, Bton Rouge, Louisiana, U.S.A.